Integration with Euler's constant and cosine

**bhuang** · April 24th 2010, 07:06 PM

Integrate

e^(2x)cosx dx

I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?

**TKHunny** · April 24th 2010, 07:08 PM

How's your "by parts"? Do it twice.

**Prove It** · April 24th 2010, 07:13 PM

Originally Posted by bhuang

Integrate

e^(2x)cosx dx

I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?

You need to use integration by parts.

$\int{u\,dv} = u\,v - \int{v\,du}$ .

So for $\int{e^{2x}\cos{x}\,dx}$

Let $u = e^{2x}$ so that $du = 2e^{2x}$ .

Let $dv = \cos{x}$ so that $v = \sin{x}$ .

Therefore $\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} - \int{2e^{2x}\sin{x}\,dx}$

$= e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx}$ .

Now you need integration by parts again.

Let $u = e^{2x}$ so that $du = 2e^{2x}$ .

Let $dv = \sin{x}$ so that $v = -\cos{x}$ .

So $e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x} - 2\left[-e^{2x}\cos{x} - \int{-2e^{2x}\cos{x}\,dx}\right]$

$= e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$ .

So we have

$\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$

$5\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x}$

$\int{e^{2x}\cos{x}\,dx} = \frac{1}{5}e^{2x}\sin{x} + \frac{2}{5}e^{2x}\cos{x} + C$ .

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